Question

find the value log(x+1)-log(x-1)=1​

Answer 1

$\log(x+1)-\log(x-1)=1$

On using the identity  $\log(a)-\log(b)=\log\left(\dfrac{a}{b}\right),$  we get,

$\log\left(\dfrac{x+1}{x-1}\right)=1$

And, since the base is 10,

$\log\left(\dfrac{x+1}{x-1}\right)=\log(10)$

Now,

$\begin{aligned}&\dfrac{x+1}{x-1}=10\\ \\ \Longrightarrow\ \ &x+1=10(x-1)\\ \\ \Longrightarrow\ \ &x+1=10x-10\\ \\ \Longrightarrow\ \ &10x-10-x-1=0\\ \\ \Longrightarrow\ \ &9x-11=0\\ \\ \Longrightarrow\ \ &x=\dfrac{11}{9}\end{aigned}$

Hence the answer is 11/9.